Explanation

(c) As shown in Figure
[image:9_image110.jpeg]
In this case refraction of the rays starting from $t_{0}$ takes place from a plane surface. So, we can use $d_{app} = \frac{d_{actual}}{\mu}$
Or $\ 3 = \frac{4}{\mu}$
Or $\mu = \frac{4}{3}$
As shown in Fig (b). In this case refraction takes place from a spherical surface. Hence, applying
$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R}$
We have,
$\frac{1}{( - 25/8)} - \frac{4/3}{- 4} = \frac{1 - 4/3}{- R}$
Or $\frac{1}{3R} = \frac{1}{3} - \frac{8}{25} = \frac{1}{75}$
$\therefore R = 25\ c$m
Now, to find the focal length we will use the lens maker formula
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$= \left( \frac{4}{3} - 1 \right)\left( \frac{1}{\infty} - \frac{1}{- 25} \right) = \frac{1}{75}$
$\therefore\ \ f =$75 cm